Electricity and Magnetism Review Faraday's Lawnebula2.deanza.edu/~lanasheridan/2B/Phys2B-Lecture21.pdfFind the total magnetic flux through the loop due to the current in the wire

Electricity and MagnetismReview

Faraday’s Law

Lana Sheridan

De Anza College

Dec 3, 2015

Overview

• Faraday’s law

• Lenz’s law

• magnetic field from a moving charge

• Gauss’s law

Reminder: Magnetic Flux

30.5 Gauss’s Law in Magnetism 917

arbitrarily shaped surface as shown in Figure 30.19. If the magnetic field at this element is B

S, the magnetic flux through the element is B

S? d A

S, where d A

S is a vec-

tor that is perpendicular to the surface and has a magnitude equal to the area dA. Therefore, the total magnetic flux FB through the surface is

FB ; 3 BS

? d AS

(30.18)

Consider the special case of a plane of area A in a uniform field BS

that makes an angle u with d A

S. The magnetic flux through the plane in this case is

FB 5 BA cos u (30.19)

If the magnetic field is parallel to the plane as in Figure 30.20a, then u 5 908 and the flux through the plane is zero. If the field is perpendicular to the plane as in Figure 30.20b, then u 5 0 and the flux through the plane is BA (the maximum value). The unit of magnetic flux is T ? m2, which is defined as a weber (Wb); 1 Wb 5 1 T ? m2.

�W Definition of magnetic flux

Figure 30.20 Magnetic flux through a plane lying in a mag-netic field.a

d

The flux through the plane is zero when the magnetic field is parallel to the plane surface.

AS

BS

b

dAS

BS

The flux through the plane is a maximum when the magnetic field is perpendicular to the plane.

Example 30.7 Magnetic Flux Through a Rectangular Loop

A rectangular loop of width a and length b is located near a long wire carrying a current I (Fig. 30.21). The distance between the wire and the closest side of the loop is c. The wire is parallel to the long side of the loop. Find the total magnetic flux through the loop due to the current in the wire.

Conceptualize As we saw in Section 30.3, the magnetic field lines due to the wire will be circles, many of which will pass through the rectangular loop. We know that the magnetic field is a function of distance r from a long wire. Therefore, the magnetic field varies over the area of the rectangular loop.

Categorize Because the magnetic field varies over the area of the loop, we must integrate over this area to find the total flux. That identifies this as an analysis problem.

S O L U T I O N

continued

br

I

c a

dr

Figure 30.21 (Example 30.7) The magnetic field due to the wire carrying a current I is not uniform over the rectangular loop.

Analyze Noting that BS

is parallel to d AS

at any point within the loop, find the magnetic flux through the rect-angular area using Equation 30.18 and incorporate Equa-tion 30.14 for the magnetic field:

FB 5 3 BS

? d AS

5 3 B dA 5 3 m0I2pr

dA

BS

u

d A S

Figure 30.19 The magnetic flux through an area element dA is B

S? d A

S5 B dA cos u, where

d AS

is a vector perpendicular to the surface.

Magnetic flux

The magnetic flux of a magnetic field through a surface A is

ΦB =∑

B · (∆A)

Units: Tm2

If the surface is a flat plane and B is uniform, that just reduces to:

ΦB = B · A

Changing flux and emf

When a magnet is at rest near a loop of wire there is no potentialdifference across the ends of the wire.

936 Chapter 31 Faraday’s Law

is moved either toward or away from it, the reading changes from zero. From these observations, we conclude that the loop detects that the magnet is moving relative to it and we relate this detection to a change in magnetic field. Therefore, it seems that a relationship exists between a current and a changing magnetic field. These results are quite remarkable because a current is set up even though no batteries are present in the circuit! We call such a current an induced current and say that it is produced by an induced emf. Now let’s describe an experiment conducted by Faraday and illustrated in Figure 31.2. A primary coil is wrapped around an iron ring and connected to a switch and a battery. A current in the coil produces a magnetic field when the switch is closed. A secondary coil also is wrapped around the ring and is connected to a sensitive ammeter. No battery is present in the secondary circuit, and the secondary coil is not electrically connected to the primary coil. Any current detected in the second-ary circuit must be induced by some external agent. Initially, you might guess that no current is ever detected in the secondary cir-cuit. Something quite amazing happens when the switch in the primary circuit is either opened or thrown closed, however. At the instant the switch is closed, the ammeter reading changes from zero momentarily and then returns to zero. At the instant the switch is opened, the ammeter changes to a reading with the opposite sign and again returns to zero. Finally, the ammeter reads zero when there is either a steady current or no current in the primary circuit. To understand what happens in this experiment, note that when the switch is closed, the current in the primary circuit produces a magnetic field that penetrates the secondary circuit. Further-more, when the switch is thrown closed, the magnetic field produced by the cur-rent in the primary circuit changes from zero to some value over some finite time, and this changing field induces a current in the secondary circuit. Notice that no current is induced in the secondary coil even when a steady current exists in the primary coil. It is a change in the current in the primary coil that induces a current in the secondary coil, not just the existence of a current. As a result of these observations, Faraday concluded that an electric current can be induced in a loop by a changing magnetic field. The induced current exists only while the magnetic field through the loop is changing. Once the magnetic field reaches a steady value, the current in the loop disappears. In effect, the loop behaves as though a source of emf were connected to it for a short time. It is cus-tomary to say that an induced emf is produced in the loop by the changing mag-netic field.

Michael FaradayBritish Physicist and Chemist (1791–1867)Faraday is often regarded as the great-est experimental scientist of the 1800s. His many contributions to the study of electricity include the invention of the electric motor, electric generator, and transformer as well as the discovery of electromagnetic induction and the laws of electrolysis. Greatly influenced by religion, he refused to work on the development of poison gas for the Brit-ish military.

© iS

tock

phot

o.co

m/S

teve

n W

ynn

Phot

ogra

phy

IN S

When a magnet is moved toward a loop of wire connected to a sensitive ammeter, the ammeter shows that a current is induced in the loop.

N S

When the magnet is held stationary, there is no induced current in the loop, even when the magnet is inside the loop.

a b

I

N S

c

When the magnet is moved away from the loop, the ammeter shows that the induced current is opposite that shown in part .a

Figure 31.1 A simple experiment showing that a current is induced in a loop when a magnet is moved toward or away from the loop.


When the north pole of the magnet is moved towards the loop, themagnetic flux increases.




© iS

tock

phot

o.co

m/S

teve

n W

ynn

Phot

ogra

phy

IN S


N S


a b

I

N S

c



A current flows clockwise in the loop.


When the north pole of the magnet is moved away from the loop,the magnetic flux decreases.




© iS

tock

phot

o.co

m/S

teve

n W

ynn

Phot

ogra

phy

IN S


N S


a b

I

N S

c



A current flows counterclockwise in the loop.

Faraday’s Law

Faraday’s Law

If a conducting loop experiences a changing magnetic flux throughthe area of the loop, an emf EF is induced in the loop that isdirectly proportional to the rate of change of the flux, ΦB withtime.

Faraday’s Law for a conducting loop:

E = −∆ΦB

∆t

Faraday’s Law

Faraday’s Law for a coil of N turns:

EF = −N∆ΦB

∆t

if ΦB is the flux through a single loop.

Changing Magnetic Flux

The magnetic flux might change for any of several reasons:

• the magnitude of B can change with time,

• the area A enclosed by the loop can change with time, or

• the angle θ between the field and the normal to the loop canchange with time.

Lenz’s Law

794 CHAPTE R 30 I N DUCTION AN D I N DUCTANCE

30-4 Lenz’s LawSoon after Faraday proposed his law of induction, Heinrich Friedrich Lenzdevised a rule for determining the direction of an induced current in a loop:

Additional examples, video, and practice available at WileyPLUS

Fig. 30-3 A coil C is located inside a solenoid S, which carries current i.

Axis

i

i

C

S

An induced current has a direction such that the magnetic field due to the currentopposes the change in the magnetic flux that induces the current.

Fig. 30-4 Lenz’s law at work.As themagnet is moved toward the loop, a currentis induced in the loop.The current producesits own magnetic field, with magnetic di-pole moment oriented so as to opposethe motion of the magnet.Thus, the in-duced current must be counterclockwise as shown.

!:

N

S

i

N

S

µ µ

The magnet's motioncreates a magneticdipole that opposesthe motion.

Furthermore, the direction of an induced emf is that of the induced current. To geta feel for Lenz’s law, let us apply it in two different but equivalent ways to Fig. 30-4,where the north pole of a magnet is being moved toward a conducting loop.

1. Opposition to Pole Movement. The approach of the magnet’s north pole inFig. 30-4 increases the magnetic flux through the loop and thereby induces acurrent in the loop. From Fig. 29-21, we know that the loop then acts as a mag-netic dipole with a south pole and a north pole, and that its magnetic dipolemoment is directed from south to north. To oppose the magnetic fluxincrease being caused by the approaching magnet, the loop’s north pole (andthus ) must face toward the approaching north pole so as to repel it (Fig.30-4). Then the curled–straight right-hand rule for (Fig. 29-21) tells us thatthe current induced in the loop must be counterclockwise in Fig. 30-4.

If we next pull the magnet away from the loop, a current will again beinduced in the loop. Now, however, the loop will have a south pole facingthe retreating north pole of the magnet, so as to oppose the retreat. Thus, theinduced current will be clockwise.

2. Opposition to Flux Change. In Fig. 30-4, with the magnet initially distant, nomagnetic flux passes through the loop. As the north pole of the magnet then

!:!:

!:

because the final current in the solenoid is zero. To find theinitial flux "B,i, we note that area A is pd2 (# 3.464 $ 10%41

4

4. The flux through each turn of coil C depends on the areaA and orientation of that turn in the solenoid’s magneticfield . Because is uniform and directed perpendicularto area A, the flux is given by Eq. 30-2 ("B # BA).

5. The magnitude B of the magnetic field in the interior of a so-lenoid depends on the solenoid’s current i and its number nof turns per unit length,according to Eq.29-23 (B # m0in).

Calculations: Because coil C consists of more than oneturn, we apply Faraday’s law in the form of Eq. 30-5(! # %N d"B/dt), where the number of turns N is 130 andd"B/dt is the rate at which the flux changes.

Because the current in the solenoid decreases at asteady rate, flux "B also decreases at a steady rate, and so wecan write d"B/dt as &"B/&t. Then, to evaluate &"B, we needthe final and initial flux values. The final flux "B, f is zero

B:B

:

m2) and the number n is 220 turns/cm, or 22 000 turns/m.Substituting Eq. 29-23 into Eq. 30-2 then leads to

Now we can write

We are interested only in magnitudes; so we ignore the mi-nus signs here and in Eq. 30-5, writing

(Answer)# 7.5 $ 10 %2 V # 75 mV.

! # N d"B

dt# (130 turns)(5.76 $ 10 %4 V)

# %5.76 $ 10 %4 Wb/s # %5.76 $ 10 %4 V.

#(0 % 1.44 $ 10 %5 Wb)

25 $ 10 %3 s

d"B

dt#

&"B

&t #

"B, f % "B,i

&t

# 1.44 $ 10 %5 Wb. $ (3.464 $ 10 %4 m2)

# (4' $ 10 %7 T (m/A)(1.5 A)(22 000 turns/m)

"B, i # BA # (!0 in)A

halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 794

Lenz’s Law

An induced current has a direction such that themagnetic field due to the current opposes thechange in the magnetic flux that induces thecurrent.

Basically, Lenz’s law let’s us interpret the minussign in the equation we write to representFaraday’s Law.

E = −−−∆ΦB

∆t

1Figure from Halliday, Resnick, Walker, 9th ed.

Lenz’s Law: Page 795 in Textbook

nears the loop with its magnetic field directed downward,the flux through the loop increases. To oppose this increase influx, the induced current i must set up its own field di-rected upward inside the loop, as shown in Fig. 30-5a; then theupward flux of field opposes the increasing downward fluxof field . The curled–straight right-hand rule of Fig. 29-21then tells us that i must be counterclockwise in Fig.30-5a.

Note carefully that the flux of always opposes thechange in the flux of , but that does not always mean that

points opposite . For example, if we next pull the mag-net away from the loop in Fig. 30-4, the flux !B from themagnet is still directed downward through the loop, but it isnow decreasing. The flux of must now be downward in-side the loop, to oppose the decrease in !B, as shown in Fig.30-5b.Thus, and are now in the same direction.

In Figs. 30-5c and d, the south pole of the magnet ap-proaches and retreats from the loop, respectively.

B:

B:

ind

B:

ind

B:

B:

ind

B:

B:

ind

B:

B:

ind

B:

ind

B:

79530-4 LE NZ’S LAWPART 3

CHECKPOINT 2

The figure shows three situations in which identical circular con-ducting loops are in uniform magnetic fields that are either in-creasing (Inc) or decreasing (Dec) in magnitude at identicalrates. In each, the dashed line coincides with a diameter. Rankthe situations according to the magnitude of the current in-duced in the loops, greatest first.

Inc

Inc

Inc

Dec

Dec

Inc

(a) (b) (c)

Fig. 30-5 The direction of the current i induced in a loop is such that the current’s magnetic field opposes the change in the magnetic field inducing i.The field is always directed opposite an increasing field and in the same direction as a decreasingfield .The curled–straight right-hand rule gives the direction of the induced current based on the direction of the induced field.B

: (b, d)

B:

(a, c)B:

indB:

B:

ind

i

Bind

B

i

BindB

i

BBind

B

Bind

i

i

Bind

B

i

BindB

i

BBind

B

Bind

i

i

Bind

B

i

BindB

i

BBindB

Bind

i

Increasing the externalfield B induces a currentwith a field Bind thatopposes the change.

The induced current creates this field, tryingto offset the change.

The fingers are in the current's direction; thethumb is in the induced field's direction.

Decreasing the externalfield B induces a currentwith a field Bind thatopposes the change.



(a) (b) (c) (d)

A


Faraday’s Law Question

The graph gives the magnitude B(t) of a uniform magnetic fieldthat exists throughout a conducting loop, with the direction of thefield perpendicular to the plane of the loop. Rank the five regionsof the graph according to the magnitude of the emf induced in theloop, greatest first.

79330-3 FARADAY’S LAW OF I N DUCTIONPART 3

As a special case of Eq. 30-1, suppose that the loop lies in a plane and that themagnetic field is perpendicular to the plane of the loop.Then we can write the dotproduct in Eq. 30-1 as B dA cos 0° ! B dA. If the magnetic field is also uniform,then B can be brought out in front of the integral sign. The remaining thengives just the area A of the loop.Thus, Eq. 30-1 reduces to

(30-2)

From Eqs. 30-1 and 30-2, we see that the SI unit for magnetic flux is the tesla–square meter, which is called the weber (abbreviated Wb):

1 weber ! 1 Wb ! 1 T " m2. (30-3)

With the notion of magnetic flux, we can state Faraday’s law in a more quantitative and useful way:

(B:

! area A, B:

uniform).#B ! BA

! dA

The magnitude of the emf ! induced in a conducting loop is equal to the rate atwhich the magnetic flux #B through that loop changes with time.

As you will see in the next section, the induced emf ! tends to oppose the fluxchange, so Faraday’s law is formally written as

(Faraday’s law), (30-4)

with the minus sign indicating that opposition. We often neglect the minus sign inEq. 30-4, seeking only the magnitude of the induced emf.

If we change the magnetic flux through a coil of N turns, an induced emf appearsin every turn and the total emf induced in the coil is the sum of these individual in-duced emfs. If the coil is tightly wound (closely packed), so that the same magnetic flux#B passes through all the turns, the total emf induced in the coil is

(coil of N turns). (30-5)

Here are the general means by which we can change the magnetic fluxthrough a coil:

1. Change the magnitude B of the magnetic field within the coil.2. Change either the total area of the coil or the portion of that area that lies

within the magnetic field (for example, by expanding the coil or sliding it intoor out of the field).

3. Change the angle between the direction of the magnetic field and the planeof the coil (for example, by rotating the coil so that field is first perpendicu-lar to the plane of the coil and then is along that plane).

B:

B:

! ! $N d#B

dt

! ! $d#B

dt

CHECKPOINT 1

The graph gives the magnitude B(t) of auniform magnetic field that existsthroughout a conducting loop,with the di-rection of the field perpendicular to theplane of the loop. Rank the five regions ofthe graph according to the magnitude ofthe emf induced in the loop, greatest first.

a b c d et

B

Sample Problem

1. Because it is located in the interior of the solenoid, coil C lieswithin the magnetic field produced by current i in the solenoid;thus,there is a magnetic flux #B through coil C.

2. Because current i decreases, flux #B also decreases.3. As #B decreases, emf ! is induced in coil C.

Induced emf in coil due to a solenoid

The long solenoid S shown (in cross section) in Fig. 30-3has 220 turns/cm and carries a current i ! 1.5 A; its diam-eter D is 3.2 cm. At its center we place a 130-turn closelypacked coil C of diameter d ! 2.1 cm. The current in thesolenoid is reduced to zero at a steady rate in 25 ms. Whatis the magnitude of the emf that is induced in coil C whilethe current in the solenoid is changing?

KEY I DEAS


Faraday’s Law

The figure shows three situations in which identical circularconducting loops are in uniform magnetic fields that are eitherincreasing (Inc) or decreasing (Dec) in magnitude at identicalrates. In each, the dashed line coincides with a diameter. Rank thesituations according to the magnitude of the current induced in theloops, greatest first.

nears the loop with its magnetic field directed downward,the flux through the loop increases. To oppose this increase influx, the induced current i must set up its own field di-rected upward inside the loop, as shown in Fig. 30-5a; then theupward flux of field opposes the increasing downward fluxof field . The curled–straight right-hand rule of Fig. 29-21then tells us that i must be counterclockwise in Fig.30-5a.

Note carefully that the flux of always opposes thechange in the flux of , but that does not always mean that

points opposite . For example, if we next pull the mag-net away from the loop in Fig. 30-4, the flux !B from themagnet is still directed downward through the loop, but it isnow decreasing. The flux of must now be downward in-side the loop, to oppose the decrease in !B, as shown in Fig.30-5b.Thus, and are now in the same direction.

In Figs. 30-5c and d, the south pole of the magnet ap-proaches and retreats from the loop, respectively.

B:

B:

ind

B:

ind

B:

B:

ind

B:

B:

ind

B:

B:

ind

B:

ind

B:

79530-4 LE NZ’S LAWPART 3

CHECKPOINT 2

The figure shows three situations in which identical circular con-ducting loops are in uniform magnetic fields that are either in-creasing (Inc) or decreasing (Dec) in magnitude at identicalrates. In each, the dashed line coincides with a diameter. Rankthe situations according to the magnitude of the current in-duced in the loops, greatest first.

Inc

Inc

Inc

Dec

Dec

Inc

(a) (b) (c)

Fig. 30-5 The direction of the current i induced in a loop is such that the current’s magnetic field opposes the change in the magnetic field inducing i.The field is always directed opposite an increasing field and in the same direction as a decreasingfield .The curled–straight right-hand rule gives the direction of the induced current based on the direction of the induced field.B

: (b, d)

B:

(a, c)B:

indB:

B:

ind

i

Bind

B

i

BindB

i

BBind

B

Bind

i

i

Bind

B

i

BindB

i

BBind

B

Bind

i

i

Bind

B

i

BindB

i

BBindB

Bind

i


The induced current creates this field, tryingto offset the change.

The fingers are in the current's direction; thethumb is in the induced field's direction.




(a) (b) (c) (d)

A


Magnetic fields from moving charges and currents

We are now moving into chapter 29.

Anything with a magnet moment creates a magnetic field.

This is a logical consequence of Newton’s third law.

Magnetic fields from moving charges

A moving charge will interact with other magnetic poles.

This is because it has a magnetic field of its own.

The field for a moving charge is given by the Biot-Savart Law:

B =µ0

4π

q v × r̂

r2

Magnetic fields from moving charges

B =µ0

4π

q v × r̂

r2

1Figure from rakeshkapoor.us.

Magnetic fields from currents

B =µ0

4π

q v × r̂

r2

We can deduce from this what the magnetic field do to the currentin a small piece of wire is.

Current is made up of moving charges!

q v = q∆s

∆t=

q

∆t∆s = I∆s

We can replace q v in the equation above.


C H A P T E R

764

M A G N E T I C F I E L D S D U E T O C U R R E N T S29

W H AT I S P H YS I C S ?29-1 One basic observation of physics is that a moving charged particleproduces a magnetic field around itself. Thus a current of moving charged parti-cles produces a magnetic field around the current. This feature of electromagnet-ism, which is the combined study of electric and magnetic effects, came as a sur-prise to the people who discovered it. Surprise or not, this feature has becomeenormously important in everyday life because it is the basis of countless electro-magnetic devices. For example, a magnetic field is produced in maglev trains andother devices used to lift heavy loads.

Our first step in this chapter is to find the magnetic field due to the current ina very small section of current-carrying wire.Then we shall find the magnetic fielddue to the entire wire for several different arrangements of the wire.

29-2 Calculating the Magnetic Field Due to a Current

Figure 29-1 shows a wire of arbitrary shape carrying a current i. We want to findthe magnetic field at a nearby point P. We first mentally divide the wire intodifferential elements ds and then define for each element a length vector thathas length ds and whose direction is the direction of the current in ds. We canthen define a differential current-length element to be i ; we wish to calculatethe field produced at P by a typical current-length element. From experimentwe find that magnetic fields, like electric fields, can be superimposed to find a netfield. Thus, we can calculate the net field at P by summing, via integration, theB

:

dB:

ds:

ds:B:

Fig. 29-1 A current-length element i produces a differential magneticfield at point P.The green (thetail of an arrow) at the dot for point Pindicates that is directed into thepage there.

dB:

!dB:

ds:

d B (into page)

Current distribution

i

P

θ ds

ids

r ˆ r

This element of current creates a magnetic field at P, into the page.

halliday_c29_764-790v2.qxd 3-12-2009 16:13 Page 764

This is another version of the Biot-Savart Law:

Bseg =µ0

4π

I ∆s × r̂

r2

where Bseg is the magnetic field from a small segment of wire, oflength ∆s.


Magnetic field around a wire segment, viewed end-on:

76529-2 CALCU LATI NG TH E MAG N ETIC F I E LD DU E TO A CU R R E NTPART 3

HALLIDAY REVISED

contributions from all the current-length elements. However, this summationis more challenging than the process associated with electric fields because ofa complexity; whereas a charge element dq producing an electric field is a scalar,a current-length element i producing a magnetic field is a vector, being theproduct of a scalar and a vector.

The magnitude of the field produced at point P at distance r by a current-length element i turns out to be

(29-1)

where u is the angle between the directions of and , a unit vector that pointsfrom ds toward P. Symbol m 0 is a constant, called the permeability constant,whose value is defined to be exactly

m 0 ! 4p " 10#7 T $ m/A ! 1.26 " 10#6 T $ m/A. (29-2)

The direction of , shown as being into the page in Fig. 29-1, is that of the crossproduct .We can therefore write Eq. 29-1 in vector form as

(Biot–Savart law). (29-3)

This vector equation and its scalar form, Eq. 29-1, are known as the law of Biotand Savart (rhymes with “Leo and bazaar”). The law, which is experimentallydeduced, is an inverse-square law. We shall use this law to calculate the net magnetic field produced at a point by various distributions of current.

Magnetic Field Due to a Current in a Long Straight WireShortly we shall use the law of Biot and Savart to prove that the magnitude of themagnetic field at a perpendicular distance R from a long (infinite) straight wirecarrying a current i is given by

(long straight wire). (29-4)

The field magnitude B in Eq. 29-4 depends only on the current and the per-pendicular distance R of the point from the wire. We shall show in our derivationthat the field lines of form concentric circles around the wire, as Fig. 29-2 showsand as the iron filings in Fig. 29-3 suggest. The increase in the spacing of the linesin Fig. 29-2 with increasing distance from the wire represents the 1/R decrease inthe magnitude of predicted by Eq. 29-4. The lengths of the two vectors in thefigure also show the 1/R decrease.

B:

B:

B:

B !% 0 i2&R

B:

dB:

!% 0

4&

i ds: ! r̂r2

ds: ! r̂dB

:

r̂ds:

dB !% 0

4&

i ds sin 'r2 ,

ds:dB

:

ds:

dB:

Fig. 29-2 The magnetic field lines pro-duced by a current in a long straight wireform concentric circles around the wire.Here the current is into the page, as indi-cated by the ".

Wire with current into the page

B

B

The magnetic field vectorat any point is tangent toa circle.

Fig. 29-3 Iron filingsthat have been sprinkledonto cardboard collect inconcentric circles whencurrent is sent through thecentral wire.The align-ment, which is alongmagnetic field lines, iscaused by the magneticfield produced by the cur-rent. (Courtesy EducationDevelopment Center)



How to determine the direction of the field lines (right-hand rule):766 CHAPTE R 29 MAG N ETIC F I E LDS DU E TO CU R R E NTS

HALLIDAY REVISED

Here is a simple right-hand rule for finding the direction of the magnetic fieldset up by a current-length element, such as a section of a long wire:

Right-hand rule: Grasp the element in your right hand with your extended thumbpointing in the direction of the current.Your fingers will then naturally curl around inthe direction of the magnetic field lines due to that element.

Fig. 29-5 Calculating the mag-netic field produced by a current i ina long straight wire.The field at Passociated with the current-length el-ement i is directed into the page,as shown.

ds:

dB:

This element of currentcreates a magnetic fieldat P, into the page.

i

θ

d B

P R

s r

ds

ˆ r

The result of applying this right-hand rule to the current in the straight wireof Fig. 29-2 is shown in a side view in Fig. 29-4a. To determine the direction of themagnetic field set up at any particular point by this current, mentally wrap yourright hand around the wire with your thumb in the direction of the current. Letyour fingertips pass through the point; their direction is then the direction of themagnetic field at that point. In the view of Fig. 29-2, at any point is tangent toa magnetic field line; in the view of Fig. 29-4, it is perpendicular to a dashed radialline connecting the point and the current.

Proof of Equation 29-4Figure 29-5, which is just like Fig. 29-1 except that now the wire is straight and ofinfinite length, illustrates the task at hand. We seek the field at point P, a per-pendicular distance R from the wire. The magnitude of the differential magneticfield produced at P by the current-length element i located a distance r from Pis given by Eq. 29-1:

The direction of in Fig. 29-5 is that of the vector —namely, directlyinto the page.

Note that at point P has this same direction for all the current-lengthelements into which the wire can be divided. Thus, we can find the magnitude ofthe magnetic field produced at P by the current-length elements in the upper halfof the infinitely long wire by integrating dB in Eq. 29-1 from 0 to !.

Now consider a current-length element in the lower half of the wire, one thatis as far below P as is above P. By Eq. 29-3, the magnetic field produced at Pby this current-length element has the same magnitude and direction as that fromelement i in Fig. 29-5. Further, the magnetic field produced by the lower halfof the wire is exactly the same as that produced by the upper half. To find themagnitude of the total magnetic field at P, we need only multiply the result ofour integration by 2.We get

(29-5)

The variables u, s, and r in this equation are not independent; Fig. 29-5 showsthat they are related by

r " 2s2 # R2

B " 2!!

0 dB "

$ 0 i2%

!!

0

sin & dsr2 .

B:

ds:

ds:

dB:

ds: ! r̂dB:

dB "$ 0

4%

i ds sin &r2 .

ds:

B:

B:

B:

Fig. 29-4 A right-hand rule gives the di-rection of the magnetic field due to a cur-rent in a wire. (a) The situation of Fig. 29-2,seen from the side.The magnetic field atany point to the left of the wire is perpen-dicular to the dashed radial line and di-rected into the page, in the direction of thefingertips, as indicated by the '. (b) If thecurrent is reversed, at any point to theleft is still perpendicular to the dashed ra-dial line but now is directed out of the page,as indicated by the dot.

B:

B:

B B

(a)

i

(b)

i The thumb is in thecurrent's direction.The fingers revealthe field vector'sdirection, which istangent to a circle.


Magnetic field from a long straight wire

The Biot-Savart Law,

Bseg =µ0

4π

I ∆s × r̂

r2

implies what the magnetic field is at a perpendicular distance Rfrom an infinitely long straight wire:

B =µ0I

2πR

(The proof requires some calculus.)

Gauss’s Law for Magnetic Fields

Gauss’s Law for magnetic fields.:∮B · dA = 0

Where the integral is taken over a closed surface A. (This is like asum over the flux through many small areas.)

We can interpret it as an assertion that magnetic monopoles donot exist.

The magnetic field has no sources or sinks.

Gauss’s Law for Magnetic Fields∮B · dA = 0

86332-3 I N DUCE D MAG N ETIC F I E LDSPART 3

CHECKPOINT 1

The figure here shows four closed surfaces with flat top and bottom faces and curvedsides.The table gives the areas A of the faces and the magnitudes B of the uniform andperpendicular magnetic fields through those faces; the units of A and B are arbitrarybut consistent. Rank the surfaces according to the magnitudes of the magnetic fluxthrough their curved sides, greatest first.

Surface Atop Btop Abot Bbot

a 2 6, outward 4 3, inwardb 2 1, inward 4 2, inwardc 2 6, inward 2 8, outwardd 2 3, outward 3 2, outward

(a) (b) (c) (d)

32-3 Induced Magnetic FieldsIn Chapter 30 you saw that a changing magnetic flux induces an electric field, andwe ended up with Faraday’s law of induction in the form

(Faraday’s law of induction). (32-2)

Here is the electric field induced along a closed loop by the changing magneticflux encircled by that loop. Because symmetry is often so powerful in physics,we should be tempted to ask whether induction can occur in the opposite sense;that is, can a changing electric flux induce a magnetic field?

The answer is that it can; furthermore, the equation governing the inductionof a magnetic field is almost symmetric with Eq. 32-2. We often call it Maxwell’slaw of induction after James Clerk Maxwell, and we write it as

(Maxwell’s law of induction). (32-3)

Here is the magnetic field induced along a closed loop by the changing electricflux !E in the region encircled by that loop.

As an example of this sort of induction, we consider the charging of a parallel-plate capacitor with circular plates. (Although we shall focus on this arrangement,

B:

! B:

! ds: " #0$0d!E

dt

!B

E:

! E:

! ds: " %d!B

dt

Fig. 32-4 The field lines for themagnetic field of a short bar mag-net.The red curves represent crosssections of closed, three-dimensionalGaussian surfaces.

B:

Surface IN

S

Surface II

B

Gauss’ law for magnetic fields holds for structures more complicated thana magnetic dipole, and it holds even if the Gaussian surface does not enclose theentire structure. Gaussian surface II near the bar magnet of Fig. 32-4 encloses nopoles, and we can easily conclude that the net magnetic flux through it is zero.Gaussian surface I is more difficult. It may seem to enclose only the north pole ofthe magnet because it encloses the label N and not the label S. However, a southpole must be associated with the lower boundary of the surface because magneticfield lines enter the surface there. (The enclosed section is like one piece of thebroken bar magnet in Fig. 32-3.) Thus, Gaussian surface I encloses a magneticdipole, and the net flux through the surface is zero.


Gauss’s Law for Magnetism Question, Ch32 # 2The figure shows a closed surface. Along the flat top face, whichhas a radius of 2.0 cm, a perpendicular magnetic field B ofmagnitude 0.30 T is directed outward. Along the flat bottom face,a magnetic flux of 0.70 mWb is directed outward. What are the(a) magnitude and(b) direction (inward or outward) of the magnetic flux through thecurved part of the surface?

884 CHAPTE R 32 MAXWE LL’S EQUATION S; MAG N ETI S M OF MATTE R

sec. 32-2 Gauss’ Law for Magnetic Fields•1 The magnetic flux through each of five faces of a die (singularof “dice”) is given by !B " #N Wb, where N (" 1 to 5) is the num-ber of spots on the face. The flux is positive (outward) for N evenand negative (inward) for N odd.What is the flux through the sixthface of the die?

•2 Figure 32-26 shows a closed surface. Alongthe flat top face, which has a radius of 2.0 cm, aperpendicular magnetic field of magnitude0.30 T is directed outward. Along the flat bot-tom face, a magnetic flux of 0.70 mWb is di-rected outward. What are the (a) magnitudeand (b) direction (inward or outward) of themagnetic flux through the curved part of thesurface?

••3 A Gaussian surface in theshape of a right circular cylinder with end caps has a radius of 12.0cm and a length of 80.0 cm. Through one end there is an inwardmagnetic flux of 25.0 mWb. At the other end there is a uniformmagnetic field of 1.60 mT, normal to the surface and directed out-ward.What are the (a) magnitude and (b) direction (inward or out-ward) of the net magnetic flux through the curved surface?

•••4 Two wires, parallel to a zaxis and a distance 4r apart,carry equal currents i in oppo-site directions, as shown in Fig.32-27. A circular cylinder of ra-dius r and length L has its axison the z axis, midway betweenthe wires. Use Gauss’ law formagnetism to derive an expres-sion for the net outward mag-netic flux through the half of the cylindrical surface above the xaxis. (Hint: Find the flux through the portion of the xz plane thatlies within the cylinder.)

sec. 32-3 Induced Magnetic Fields•5 The induced magnetic field at radial distance 6.0 mmfrom the central axis of a circular parallel-plate capacitor is 2.0 $10%7 T. The plates have radius 3.0 mm. At what rate is theelectric field between the plates changing?

•6 A capacitor with square plates of edgelength L is being discharged by a current of 0.75A. Figure 32-28 is a head-on view of one of theplates from inside the capacitor. A dashed rec-tangular path is shown. If L " 12 cm, W " 4.0cm, and H " 2.0 cm, what is the value of ! around the dashed path?

••7 Uniform electric flux. Figure 32-29shows a circular region of radius R " 3.00 cmin which a uniform electric flux is directed out of the plane of the

B:

! ds:

dE:

/dt

SSM

ILWSSM

B:

page. The total electric flux through the region isgiven by !E " (3.00 mV & m/s)t, where t is in sec-onds. What is the magnitude of the magneticfield that is induced at radial distances (a) 2.00cm and (b) 5.00 cm?

••8 Nonuniform electric flux. Figure 32-29 shows a circular region of radius R " 3.00cm in which an electric flux is directed out ofthe plane of the page. The flux encircled by aconcentric circle of radius r is given by !E,enc "(0.600 V & m/s)(r/R)t, where r ' R and t is in seconds. What is themagnitude of the induced magnetic field at radial distances (a) 2.00cm and (b) 5.00 cm?

••9 Uniform electric field. In Fig. 32-29, a uniform electric fieldis directed out of the page within a circular region of radius R " 3.00cm. The field magnitude is given by E " (4.50 $ 10%3 V/m & s)t,where t is in seconds.What is the magnitude of the induced magneticfield at radial distances (a) 2.00 cm and (b) 5.00 cm?

••10 Nonuniform electric field. In Fig. 32-29, an electric fieldis directed out of the page within a circular region of radius R "3.00 cm.The field magnitude is E " (0.500 V/m & s)(1 % r/R)t, where tis in seconds and r is the radial distance (r ' R). What is the magni-tude of the induced magnetic field at radial distances (a) 2.00 cm and(b) 5.00 cm?

••11 Suppose that a parallel-plate capacitor has circular plateswith radius R " 30 mm and a plate separation of 5.0 mm. Supposealso that a sinusoidal potential difference with a maximum value of150 V and a frequency of 60 Hz is applied across the plates; that is,

V " (150 V) sin[2p(60 Hz)t].

(a) Find Bmax(R), the maximum value of the induced magnetic fieldthat occurs at r " R. (b) Plot Bmax(r) for 0 ( r ( 10 cm.

••12 A parallel-plate capacitor with circular plates of radius 40mm is being discharged by a current of 6.0 A.At what radius (a) in-side and (b) outside the capacitor gap is the magnitude of the in-duced magnetic field equal to 75% of its maximum value? (c) Whatis that maximum value?

sec. 32-4 Displacement Current•13 At what rate must the potential difference between theplates of a parallel-plate capacitor with a 2.0 mF capacitance bechanged to produce a displacement current of 1.5 A?

•14 A parallel-plate capacitor with circular plates of radius R isbeing charged. Show that the magnitude of the current density ofthe displacement current is Jd " )0(dE/dt) for r ' R.

•15 Prove that the displacement current in a parallel-platecapacitor of capacitance C can be written as id C(dV/dt), whereV is the potential difference between the plates.

•16 A parallel-plate capacitor with circular plates of radius 0.10m is being discharged.A circular loop of radius 0.20 m is concentric

"SSM

Fig. 32-26Problem 2.

B

Fig. 32-27 Problem 4.

x

y

r–r

Wire 1

–2r 2r

Wire 2

Fig. 32-28Problem 6.

LH

W

L

Fig. 32-29Problems 7 to

10 and 19 to 22.

R

Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign

SSM Worked-out solution available in Student Solutions Manual

• – ••• Number of dots indicates level of problem difficulty

Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com

WWW Worked-out solution is at

ILW Interactive solution is at http://www.wiley.com/college/halliday


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Summary

• Faraday’s law

• Lenz’s law

• magnetic field from a moving charge

• Guass’s law

HomeworkStudy!

Documents

Electricity and Magnetism Review Faraday's Lawnebula2.deanza.edu/~lanasheridan/2B/Phys2B-Lecture21.pdfFind the total magnetic flux through the loop due to the current in the wire